How do I balance these reactions? Also, how can I write the net ionic reaction for the second one?

1)

#color(red)(?)"NH"_4"NO"_3(s) stackrel(Delta" ")(->) color(red)(?)"N"_2"O"(g) + color(red)(?)"H"_2"O"(g)#

This is a thermal decomposition reaction, with gases formed, so it likely involves heating the pure solid. That's why I put #Delta#, which means add heat.

Note that the nitrogens are already balanced. There is also an odd number of oxygens. That tells you that the coefficient for #"N"_2"O"# is just #1#.

Therefore, we only touch #"H"_2"O"# and give it a coefficient of #2#. That indeed gives #4# equivalents of #"H"# on both sides, and #3# equivalents of #"O"# on both sides.

#color(blue)("NH"_4"NO"_3(s) stackrel(Delta" ")(->) "N"_2"O"(g) + \mathbf(2)"H"_2"O"(g))#

2)

#color(red)(?)"Cu"(s) + color(red)(?)"AgNO"_3(aq) -> color(red)(?)"Ag"(s) + color(red)(?)"Cu"("NO"_3)_2(aq)#

We know that there are two #"NO"_3^(-)# polyatomic ions on the products side, so there must be #2# equivalents of #"AgNO"_3# to balance it (you can't change the number of #"NO"_3# on the reactants side without touching #"Ag"#).

Then, we have unbalanced the #"Ag"#, requiring #2# equivalents of #"Ag"(s)# on the products side as well to rebalance #"Ag"#.

#color(blue)("Cu"(s) + \mathbf(2)"AgNO"_3(aq) -> \mathbf(2)"Ag"(s) + "Cu"("NO"_3)_2(aq))#

This is the complete molecular equation of a redox reaction. You can see that if you wrote this as a net ionic equation, you would get:

#color(green)("Cu"(s) + \mathbf(2)"Ag"^(+)(aq) -> \mathbf(2)"Ag"(s) + "Cu"^(2+)(aq))#

Suppose we have a copper anode and a silver cathode in an galvanic cell.

In this reaction, what we have happening is that aqueous silver cation is getting reduced in solution (#"Ag"^(+)(aq) + e^(-) -> "Ag"(s)# is the half-reaction) due to the flow of electrons through the conductive wire rightwards, and thus forming more solid silver at the cathode (right beaker).

The solid copper is getting oxidized (#"Cu"(s) -> "Cu"^(2+) + 2e^(-)# is the half reaction), sending electrons through the conductive wire rightwards, thus forming a cation in solution, at the anode (left beaker).